= Solution
Put $\tau=a\cdot(b\times c)$. The <vector triple product identity> gives
$$
(b\times c)\times(c\times a)
=[(b\times c)\cdot a]c-[(b\times c)\cdot c]a=\tau c.
$$
Cyclic invariance of the <scalar triple product> gives $(a\times b)\cdot c=\tau$. Hence the <squared scalar triple product from cyclic cross products> is
$$
\boxed{(a\times b)\cdot[(b\times c)\times(c\times a)]=[a\cdot(b\times c)]^2.}
$$
If an expansion entirely in pairwise <dot products> is desired, expand the <Gram determinant> of $a,b,c$:
$$
\boxed{|a|^2|b|^2|c|^2+2(a\cdot b)(b\cdot c)(c\cdot a)
-|a|^2(b\cdot c)^2-|b|^2(c\cdot a)^2-|c|^2(a\cdot b)^2.}
$$
Both expressions remain valid for degenerate, including linearly dependent, vectors.
Back to article page
